Trying to solve the recurrence relation by comparing 3 cases of master theorem

Question

I am trying to understand how the master theorem is invoked on the following recurrence relation:

$$ T(n) = \sqrt{6006} T(n/2) + n^{\sqrt{6006}}. $$

So basically, I found the following source where they have Master theorem definition defined.

I found the similar solution (See the Link #1 in the comment below) where they are trying to solve the same problem (Problem #c). I

In the solution, they have mentioned the following :

We have :

n^0.5log_b(a) = n^{log_b(6006)1/2}

= n^{0.5log₂(6006)1/2}

and so on and so forth. I am not able to understand how they compared the results with the given equation.

Answer 1

You determine $a$, $b$, and $f$ using pattern matching.

The Master theorem applies for recurrences of the form

$\qquad\displaystyle T(n) = a T(n/b) + f(n)$;

you have

$\qquad\displaystyle T(n) = \sqrt{6006} \cdot T(n/2) + n^{\sqrt{6006}}$.

Just by comparing the two formulae, you determine that

$\qquad\begin{align*} a &= \sqrt{6006},\\ b &= 2, \quad\text{and} \\ f &= n \mapsto n^{\sqrt{6006}}. \end{align*}$

Now you go through the three cases and check which applies (if any). See also here.